Question

what is the main product when pa-233 undergoes alpha decay?
u-233
u-235
ac-229
ra-233
question 21 (0.0862 points)
a sample of 0.225 mole of gas has a volume of 748 ml at 28°c. calculate the pressure in atm of this gas. (r= 0.0821 l·atm / mol·k). give the number with the correct significant figures and unit.
question 22 (0.0862 points)
which solution below has the lowest concentration of hydroxide ions?

Explanation:

Question 1 (Alpha Decay of Pa-233)

Step 1: Recall Alpha Decay Rules

In alpha decay, a nucleus emits an alpha particle ($\ce{^4_2He}$), so the mass number decreases by 4 and the atomic number decreases by 2. Protactinium (Pa) has an atomic number of 91.

Step 2: Calculate New Mass and Atomic Numbers

• Mass number of Pa-233: 233. After alpha decay, new mass number = \( 233 - 4 = 229 \).
• Atomic number of Pa: 91. After alpha decay, new atomic number = \( 91 - 2 = 89 \).
• Element with atomic number 89 is Actinium (Ac). So the product is Ac-229.

Step 1: Convert Units

• Volume: \( 748 \, \text{mL} = 0.748 \, \text{L} \) (since \( 1 \, \text{L} = 1000 \, \text{mL} \)).
• Temperature: \( T = 28^\circ\text{C} + 273.15 = 301.15 \, \text{K} \).

Step 2: Apply Ideal Gas Law (\( PV = nRT \))

We need to solve for \( P \): \( P = \frac{nRT}{V} \).

• \( n = 0.225 \, \text{mol} \), \( R = 0.0821 \, \text{L·atm/mol·K} \), \( T = 301.15 \, \text{K} \), \( V = 0.748 \, \text{L} \).
• Substitute values: \( P = \frac{0.225 \times 0.0821 \times 301.15}{0.748} \).
• Calculate numerator: \( 0.225 \times 0.0821 = 0.0184725 \); \( 0.0184725 \times 301.15 \approx 5.563 \).
• Divide by volume: \( P \approx \frac{5.563}{0.748} \approx 7.44 \, \text{atm} \).

Answer:

C. Ac-229

Question 21 (Gas Law Calculation)